Answer the following questions : (i) The conductivity of 0.20 M solution of KCl is 2.48 × 10^-2 S cm^-1. Calculate its molar conductivity and degree of dissociation (α). [Given : λ°(K^+) = 73.5 S cm² mol^-1, λ°(Cl^-) = 76.5 S cm² mol^-1] (ii) Calculate Δ_rG° of the following cell : Mg(s) + Cu²+(aq) → Mg²+(aq) + Cu(s) [Given : E°Mg²+/Mg = – 2.37 V, E°Cu²+/Cu = + 0.34 V, 1 F = 96500 C mol^-1] (iii) What type of cell is mercury cell ? Why is it more advantageous than dry cell ?

ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2025: question 33(b)

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerElectrochemistry

Answer the following questions :

(i)
The conductivity of 0.20 M solution of KCl is $2.48 \times 1 0^{- 2}$ S cm $^{- 1}$ . Calculate its molar conductivity and degree of dissociation ( $α$ ).
[Given : $λ_{(K^{+})}^{\circ}$ = 73.5 S cm $^{2}$ mol $^{- 1}$ , $λ_{(C l^{-})}^{\circ}$ = 76.5 S cm $^{2}$ mol $^{- 1}$ ]
2 marks
(ii)
Calculate $Δ_{r} G^{\circ}$ of the following cell :
$M g (s) + C u^{2 +} (a q) \to M g^{2 +} (a q) + C u (s)$
[Given : $E_{M g^{2 +} / M g}^{\circ}$ = – 2.37 V, $E_{C u^{2 +} / C u}^{\circ}$ = + 0.34 V, 1 F = 96500 C mol $^{- 1}$ ]
2 marks
(iii)
What type of cell is mercury cell ? Why is it more advantageous than dry cell ?
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Use $Λ_{m} = κ \times 1000/ C$ and Kohlrausch's law for $Λ_{m}^{\circ}$ .

Hint 2

α is the ratio of molar conductivity at the given concentration to limiting molar conductivity.

Hint 3

For $Δ_{r} G^{\circ}$ , find $E_{ce l l}^{\circ}$ with Mg as the anode, then use $- n F E^{\circ}$ with n = 2.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Molar conductivity:
$Λ_{m} = \frac{κ \times 1000}{C} = \frac{2.48 \times 1 0 ^{- 2} \times 1000}{0.2}$
½ mark
$Λ_{m} = 124 S cm^{2} mol^{- 1}$
½ mark
Limiting molar conductivity:
$Λ_{m}^{\circ} = λ_{K^{+}}^{\circ} + λ_{C l^{-}}^{\circ} = 73.5 + 76.5 = 150 S cm^{2} mol^{- 1}$
½ mark
Degree of dissociation:
$α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{124}{150} = 0.827$
½ mark
(ii) Mg is oxidised (anode) and $C u^{2 +}$ is reduced (cathode):
$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = 0.34 - (- 2.37)$
½ mark
$E_{ce l l}^{\circ} = 2.71 V$
½ mark
With $n = 2$ :
$Δ_{r} G^{\circ} = - n F E_{ce l l}^{\circ} = - 2 \times 96500 \times 2.71$
½ mark
$Δ_{r} G^{\circ} = - 523030 J mol^{- 1} = - 523.03 kJ mol^{- 1}$
½ mark
(iii) The mercury cell is a primary cell.
½ mark
It is better than a dry cell because it gives a constant potential (about 1.35 V) throughout its life. This is because the overall reaction involves no ions whose concentration changes. It also lasts longer.
½ mark

Answer: (i) $Λ_{m} = 124$ S cm² mol⁻¹, α = 0.827; (ii) $E_{ce l l}^{\circ} = 2.71$ V, $Δ_{r} G^{\circ} = - 523030$ J mol⁻¹ (−523.03 kJ mol⁻¹); (iii) primary cell; gives constant voltage throughout its life

Where the marks are

Formula $Λ_{m} = κ \times 1000/ C$ with substitution (½ mark)
$Λ_{m} = 124$ S cm² mol⁻¹ (½ mark)
$Λ_{m}^{\circ} = 150$ S cm² mol⁻¹ (½ mark)
α = 124/150 = 0.827 (½ mark)
$E_{ce l l}^{\circ} = E_{c}^{\circ} - E_{a}^{\circ}$ with substitution (½ mark)
$E_{ce l l}^{\circ} = 2.71$ V (½ mark)
$Δ_{r} G^{\circ} = - n F E^{\circ}$ with n = 2 substituted (½ mark)
$Δ_{r} G^{\circ} = - 523030$ J mol⁻¹ (½ mark)
Mercury cell is a primary cell (½ mark)
Constant potential throughout use / longer life (½ mark)

Revise the chapter

Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
Topic: Conductivity, Molar Conductivity and Kohlrausch's Law

Practise this chapterBoard-style questions from Electrochemistry, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 56/7/1: every question, by section.

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CBSE Class 12 Chemistry 2025: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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